Equation of motion - choosing directions for the coordinate axis

It is up to you to decide which coordinate system you'd like to use.

In your case, you have chosen the downward direction of the y-axis to be the positive direction. Therefore, the sign of the $mg$ will be positive.

Since the $mg$ force is acting downwards, the term will be positive. The force which is acting upwards will be negative.

$mddot{y}=mg-{frac{rho A C}{2}dot{y}^2}$

Your diagram however, hints that you should use the upward direction as the positive direction. In this case, the signs of the quantities will flip.

$mddot{y}={frac{rho A C}{2}dot{y}^2} - mg$

I believe that the original equation as presented in the OP is ill-formed.

A correct vector equation should not be affected by an arbitrary choice as to the positive direction for the vectors.

In the equation given, the LHS will change sign under such a change in the positive direction. A landing space ship that is slowing to a stop will have a numerically positive upward acceleration, or a numerically negative downward acceleration.

Likewise the acceleration of gravity (second term, RHS) will numerically change in sign. If up is positive, $g=-9.8$: if down is positive, $g=9.8$

However, the drag term, with the square of the velocity, is always non-negative. It does not switch signs, and the equation fails.

The inclusion of a unit vector in the drag term, with a direction opposed to the velocity, would make the equation correct...

The correct form of the equation is

$$ mddot{y}=mg-{frac{rho A C}{2}|dot{y}|,dot{y}} $$

or

$$ begin{cases} mddot{y}=mg-{frac{rho A C}{2}dot{y}^2} & dot{y}>0 mddot{y}=mg+{frac{rho A C}{2}dot{y}^2} & dot{y}<0 end{cases} $$

This is because air resistance must oppose the motion $dot{y}$.